Question

If $a$ and $b$ are roots of the equation $x^2-lx+m=0$ and $\frac{a}{3},3b$ are the roots of the equation $x^2-nx+m=0$ then $m$ equals

• A. $(3n-l)(3l-n)$
• B. $\frac{(3n+1)(3l+n)}{64}$
• C. $\frac{(l-3n)(3l-n)}{64}$
• D. None of these

Answer 1

Answer: (D)

None of these

As $a,b$ are roots of $x^2-lx+m=0$

$\therefore a+b=l,ab=m$

and $\frac{a}{3},3b$ are roots of the equation $x^2-nx+m=0$

$\therefore \frac{a}{3}+3b=n,ab=m$ now $a+b=l,a+9b=3n$

$\Rightarrow b=\frac{3n-l}{8}\therefore a=\frac{3}{8}(3l-n)$

now $m=ab=\frac{3}{64}(3n-l)(3l-n)$